Question

week 7
find all zeros of this polynomial (f(x)=x^{3}+10x^{2}+13x - 24) if (x = - 8) is a zero. what are the rest?
zeros: (x=-8,x=), (x=)

Explanation:

Step1: Use synthetic division

Since \(x = - 8\) is a zero, we perform synthetic - division on \(f(x)=x^{3}+10x^{2}+13x - 24\) with \(-8\).
The coefficients are \(1,10,13,-24\).
Bring down the first coefficient \(1\):
\[

$$\begin{array}{c|cccc} -8&1&10&13&-24\\ && - 8&-16&24\\ \hline &1&2&-3&0 \end{array}$$

\]
The quotient is \(x^{2}+2x - 3\).

Step2: Factor the quotient

We factor the quadratic \(x^{2}+2x - 3\).
We need to find two numbers that multiply to \(-3\) and add up to \(2\). The numbers are \(3\) and \(-1\).
So \(x^{2}+2x - 3=(x + 3)(x - 1)\).

Step3: Find the remaining zeros

Set \((x + 3)(x - 1)=0\).
If \(x+3 = 0\), then \(x=-3\); if \(x - 1=0\), then \(x = 1\).

Answer:

\(x=-3,x = 1\)